Prove or disprove:

*Proof.* We can compute using the identities we have derived in the previous exercises of this section,

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Stumbling Robot

A Fraction of a Dot
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Tag: Cross Products

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Prove or disprove a given vector formula

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Prove an identity relating scalar triple products of vectors *A,B,C*

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Compute the cross product of given vectors in terms of the unit coordinate vectors

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Prove some properties of the scalar triple product

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Prove that there is exactly one vector satisfying given conditions

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Find a vector that is equal to the cross product of two given vectors

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Prove some facts about orthogonal unit vectors

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Prove that the norm of the cross product is the product of the norms if and only if *A* and *B* are orthogonal

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Express the cross product of given vectors in terms of the unit coordinate vectors

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Use the cross product to compute the area of triangles with given vertices

Prove or disprove:

*Proof.* We can compute using the identities we have derived in the previous exercises of this section,

Prove the following identity holds for vectors .

*Proof.* From a previous exercise (Section 13.14, Exercise #9(d)), we know

So, with in place of and in place of in this formula we have

Let such that

In terms of the unit coordinate vectors compute the cross product

Using part (a) of the previous exercise and equation (13.10) on page 490 of Apostol () we compute

Then we use the other given relations

Which all implies

Use the properties of the cross product and the dot product to prove the following properties of the scalar triple product.

- .
- .
- .
- .

*Proof.*We havesince for any vectors (in this case and

*Proof.*Using part (a), we have*Proof.*We have*Proof.*We have

Prove that there is one and only one vector satisfying the equations

where and is orthogonal to in .

*Proof.* First, we show uniqueness. If is any vector such that and and is another vector such that and then we have

But, by a previous exercise (Section 13.11, Exercise #8) we know these conditions imply . Hence, there is at most one such .

Now, for existence. Let . Since are orthogonal we know ,

From we then have

Since , we know at least one of the . Without loss of generality, assume . From the second and third equations we have

This implies

Hence, can take any value. The vectors such that are then of the form

Then,

Since (since at least one of the , we have that such a vector always exists

Consider the vectors

- Find satisfying . How many such solutions are there?
- Find such that and . How many such solutions are there?

- Let . For we must have
Therefore, we have the three equations,

From the first equation we have . From the second equation we then have . Since any value of then satisfies the third equation we have that is arbitrary. Letting we then have and . Hence, a solution is

There are infinitely many solutions since we can take any value for to obtain another solution.

- From part (a) we know that the vectors such that are of the form
for any value of . Then,

From part (a) we then have and . Hence, is the only solution.

Let be orthogonal unit vectors.

- Prove that the vectors form an orthonormal basis for .
- Prove that the vector has unit length.
- Show the geometric relation between and obtain the relations
- Prove the relations in part (c) algebraically.

*Proof.*By Theorem 13.13 (page 484 of Apostol) we know that if and are independent then so is the set . We also know by Theorem 13.12 (page 483) that is orthogonal to both and . Since this is a set of three independent vectors in , it is a basis. Then, if each have length 1 and are orthogonal we haveHence, has length 1 as well. Thus, is form an orthonormal basis

*Proof.*From part (a) we know that and each have length 1 and are orthogonal. Thus,-
**Incomplete.** *Proof.*Since are orthogonal, we know every vector orthogonal to two of them is a scalar multiple of the third. Thus, is a scalar multiple of . Further, since all have length 1If we adopt a right hand coordinate system then . So, . Similarly,

Given vectors prove that

if and only if and are orthogonal.

*Proof.*

From Theorem 13.12(f) (page 483 of Apostol) we know

But if and only if and are orthogonal (from the definition of orthogonality). Thus, if and only if and are orthogonal

Let

Find in terms of .

First, we have

So, the cross product is given by

For each of the following sets of points use the cross product to compute the area of the triangle with vertices .

- ;
- ;
- .

- The area of the triangle with vertices is given by
- The area of the triangle with vertices is given by
- The area of the triangle with vertices is given by